Python 3 lambda error: The truth value of a Series is ambiguous

Question

I am getting this error: The truth value of a Series is ambiguous in my lambda function. I know that here is a very comprehensive explanation around this error but I don't think this relates to my issue: Truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()

Basically I am trying to determine via lambda whether OpenBal is the same from one month to the next within the same AccountID and give me a '1' if it is the same (e.g. for OpenBal=101 below). Obviously the first record should give me a NaN. (P.S. thanks @jdehesa for your answers in my other post).

This demonstrates my problem:

import pandas as pd
df = pd.DataFrame({'AccountID': [1,1,1,1,2,2,2,2,2],
                   'RefMonth':    [1,2,3,4,1,2,3,4,5],
                   'OpenBal':    [100,101,101,103,200,201,202,203,204]})
SameBal = df.groupby('AccountID').apply(lambda g: 1 if g['OpenBal'].diff() == 0 else 0)
df['SameBal'] = SameBal.sortlevel(1).values

I am just wondering, are nested lambdas appropriate for the problem. That is something of the general form df.apply(lambda g:g.apply())? — demongolem, May 27 '20 at 16:43

Dillon · Answer 1 · 2018-06-20T07:28:39.763

1 if g['OpenBal'].diff() == 0 is not working. This is not how the pd.Series() object can operate

You need to create a suitable method:

def convert(a):
    return np.array([1 if i==0 else np.nan if pd.isnull(i) else 0 for i in a])

This will solve your The truth value of a Series is ambiguous error

SameBal = df.groupby('AccountID').apply(lambda g: pd.Series(data=convert(g['OpenBal'].diff().values), index=g['RefMonth']))
SameBal.name = 'SameBal'

SameBal 
Out[]:
AccountID  RefMonth
1          1           NaN
           2           0.0
           3           1.0
           4           0.0
2          1           NaN
           2           0.0
           3           0.0
           4           0.0
           5           0.0

df.merge(SameBal.reset_index())
Out[]:
   AccountID  OpenBal  RefMonth  SameBal
0          1      100         1      NaN
1          1      101         2      0.0
2          1      101         3      1.0
3          1      103         4      0.0
4          2      200         1      NaN
5          2      201         2      0.0
6          2      202         3      0.0
7          2      203         4      0.0
8          2      204         5      0.0

jpp · Accepted Answer · 2018-06-20T09:32:18.437

Your error correctly indicates you can't check the truthness of a series. But custom anonymous functions are not necessary for this task.

Using groupby + transform with pd.Series.diff:

import pandas as pd

df = pd.DataFrame({'AccountID': [1,1,1,1,2,2,2,2,2],
                   'RefMonth':    [1,2,3,4,1,2,3,4,5],
                   'OpenBal':    [100,101,101,103,200,201,202,203,204]})

df['A'] = (df.groupby('AccountID')['OpenBal'].transform(pd.Series.diff)==0).astype(int)

print(df)

   AccountID  OpenBal  RefMonth   A
0          1      100         1   0
1          1      101         2   0
2          1      101         3   1
3          1      103         4   0
4          2      200         1   0
5          2      201         2   0
6          2      202         3   0
7          2      203         4   0
8          2      204         5   0

If you need NaN for the first row of each group:

g = df.groupby('AccountID')['OpenBal'].transform(pd.Series.diff)
df['A'] = (g == 0).astype(int)
df.loc[g.isnull(), 'A'] = np.nan

print(df)

   AccountID  OpenBal  RefMonth    A
0          1      100         1  NaN
1          1      101         2  0.0
2          1      101         3  1.0
3          1      103         4  0.0
4          2      200         1  NaN
5          2      201         2  0.0
6          2      202         3  0.0
7          2      203         4  0.0
8          2      204         5  0.0

Nicely done! @Dillon, I might need your custom function for a similar task, so thanks also... — GivenX, Jun 20 '18 at 09:24
@Givenx This answer does not show NaN as per the question: 'Obviously the first record should give me a NaN' — Dillon, Jun 20 '18 at 09:28
@Dillon, Sure, I've updated with a solution for this scenario. — jpp, Jun 20 '18 at 09:32

Python 3 lambda error: The truth value of a Series is ambiguous

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